Consider the curve given by the equation $$y^3-xy=2$$. It can be shown that $$\frac{dy}{dx}$$ = $$\frac{y}{3y^2-x}$$. Write an equation for the line tangent to the curve at the point (-1,1).

Cole S.

Answer:

The line of tangency will be in the form of $$y=mx+b$$. Our $$m$$ is equivalent to $$\frac{dy}{dx}$$ given our point. Therefore, $$m$$=$$\frac{1}{4}$$. Further, $$y=mx+b$$ is then $$1=\frac{1}{4}(-1)+b$$. When we solve for $$b$$, we find it to be $$\frac{5}{4}$$. There fore our line of tangency at the point (-1,1) is $$y=\frac{1}{4}x+\frac{5}{4}$$.

Algebra

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Question:

The length of a rectangle is 1 meter more than two times its width. If the area of the rectangle is 36 square meters, what are its dimensions?

Cole S.

Answer:

Width: 4, Length: 9

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